But this still confuses me. Let me explain what i thought was happening without the capacitor in the first place...a simple voltage divider with a A.C input. The a.c input doesnt show in the picture
The voltage divider(r1/r2) set up a voltage at re, thus creating an Ic(q) independent of Beta.
Now this is how i believed the A.C input signal created an A.C output signal...i had two theories...
1)The A.C input was superimposed on r2, thus creating an a.c voltage at R2..voltage changes at R2 created a voltage change to Re.So we had created an A.C voltage on Re, thus creating an A.c current in the emitter/collector loop.
2)The A.C input would travel through the base-emitter junction and superimpose on Re, creating an A.C voltage on Re, thus creating an A.C current in the emitter/collector loop.
Now with the input A.C shorted at the emitter how can the A.C input being aplied only at the base-emitter junction create an output A.C.